Vector space/Dual basis/Base change/Fact/Proof

Proof

We have

${}{\begin{aligned}{\left(\sum _{i=1}^{n}b_{ij}v_{i}^{*}\right)}(w_{r})&={\left(\sum _{i=1}^{n}b_{ij}v_{i}^{*}\right)}{\left(\sum _{k=1}^{n}a_{kr}v_{k}\right)}\\&=\sum _{1\leq i,k\leq n}b_{ij}a_{kr}v_{i}^{*}(v_{k})\\&=\sum _{i=1}^{n}b_{ij}a_{ir}.\end{aligned}}$

Here, we have the "product“ of the ${}j$-th column of ${}B$ and the ${}r$-th column of ${}A$, which is also the product of the ${}j$-th row of ${}{B^{\text{tr}}}=A^{-1}$ and the ${}r$-te column of ${}A$. For ${}r=j$, this is ${}1$, and for ${}r\neq j$, this is ${}0$. Therefore, the given linear form coincides with ${}w_{j}^{*}$.

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